Pulley rotation 
In the given arrangement, the friction in between the string and the pulley is enough to avoid slipping and the string is light. On the basis of these assumptions, would the tension in the upper part of the string (ie. the arc BC)  be 0? Because AB is in equilibrium as the frictional force equals the tension in AB acting on AB by the mass m2, the same for DC, hence there is no force on the part BC and thus the tension should be 0.
Am I correct?, any help will be appreciated 
 A: The only way that friction can appear is for there to be tension, since it is tension that will give rise to the normal force needed for friction.
Now if we note that the friction must result in a difference in tension between the left and right strings (if the masses are different) then there will be a continuous change in tension. The normal force at every point is proportional to the local tension, and in the limit where the friction is "just enough" this means that there will be an exponential decay in the tension from the high end to the low end.
The diagram that helps explain this is the following:

If you consider just a bit of string that is touching a small arc of the pulley, then there is a normal force $F_n$ due to the tension in the string that is proportional to the local tension $T$ and the angle $d\theta$, and that gives rise to a differential friction $dF = \mu F_n$. As a consequence, the tension $T_2$ will be larger than $T_1$ by that amount: $T_2 = T_1 + dF$. When the angle $d\theta$ becomes very small, you can consider $T_1 = T_2 = T$ for the purpose of computing the normal force, $F_n = Td\theta$.
See if you can derive the rest from there.
A: If the pulley is free to rotate and perfectly frictionless then the tension in the string is the same throughout, from A to D.  So the tension in BC is not zero.
You are correct in your comment stating that the net force on BC is zero, because the tensions in AB pulling right and in CD pulling left are equal.  However, the net force acting on section BC is not the same thing as the tension in section BC.  It can be in static equilibrium while still being in tension.    
